Proof That 0/0 = 1 Based on x^0 Equaling 1?

Date: 02/02/2006 at 04:12:53
From: Joseph
Subject: x^0 vs. 0/0
You wrote that the reason x^0 = 1 is because of the laws of exponents:
(3^4)/(3^4) = 3^(4-4) = 3^0.
You also wrote that 0^0 = 1 while maintaining that 0/0 doesn't make
sense. If the proof for any x^0 is 1 is through exponents, then how
do we prove 0^0 = 1?
(0^3)/(0^3) = 0^(3-3) = 0^0 but (0^3) = 0 so: 0/0 = 1
I am specifically targeting Euler's argument to make all x^0 = 1.
Does he have a different method of proof besides exponents?

Date: 02/02/2006 at 09:38:24
From: Doctor Peterson
Subject: Re: x^0 vs. 0/0
Hi, Joseph.
It sounds like you may not have seen our FAQ on this topic:
0 to 0 Power
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
The fact is, we DON'T prove that 0^0 is 1!
When x is not 0, x^0 is equal to 1, because that definition is
consistent with the rules for exponents, as you mentioned:
x^0 = x^(1-1) = x^1 / x^1 = x/x = 1
These rules don't work when x = 0, however. In order to choose a
reasonable definition for this, we have to consider limits: what
happens to x^y when both x and y approach zero? It turns out that the
answer depends on how they approach zero: if x goes to zero first,
then x^y = 0^y = 0 for all y, so the limit is 0; but if y goes to zero
first, then x^y = x^0 = 1 for all x except zero, and the limit is 1.
We call such a limit "indeterminate", meaning that as it stands there
is no way to choose one correct value.
That is all we can do in general. However, 0^0 is an unusual
indeterminate case, in that for most purposes the value 1 is
appropriate. When working with those formulas, we arbitrarily define
0^0 as equal to 1, so that the formulas work neatly. This is not
something we prove, but just a definition that works in these cases.
The FAQ gives some examples of the reason for this choice.
Note that although 0^0 is therefore defined as 1 for use in certain
formulas, it is still true that 0/0 as a limit is still indeterminate,
and must be resolved specially in each problem in which it occurs.
Also, we have to be aware that 0^0 can't be treated as 1 in every
situation, as illustrated by your "proof" that 0/0 = 1.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/